Question

solve the following mole related problems
a) how many moles of rubidium phosphate are present in 97.45 grams?

Explanation:

Step1: Determine the molar mass of rubidium phosphate ($Rb_3PO_4$)

The molar mass of $Rb$ is $85.47\ g/mol$, of $P$ is $30.97\ g/mol$, and of $O$ is $16.00\ g/mol$.
$$M = 3\times85.47+30.97 + 4\times16.00$$
$$M=256.41+30.97+64.00$$
$$M = 351.38\ g/mol$$

Step2: Calculate the number of moles

Use the formula $n=\frac{m}{M}$, where $m = 97.45\ g$ and $M = 351.38\ g/mol$.
$$n=\frac{97.45}{351.38}\approx0.277\ mol$$

Answer:

Approximately $0.277$ moles.